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| :'''consider a particle with charge e moving under three dimensional isotropic harmonic potential '''l
| | The Hamiltonian of the system is |
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| <math>V(r)=\frac{1}{2}m{\omega }^2{r}^2</math>'''in an electric field''' <math>E=E_{0}(x)</math> '''Find the eigen states and eigen values of the patricle''' | | <math>H=\frac{p^2}{2m}+\tfrac{1}{2}m\omega^2r^2-eE_{0}x.</math> |
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| the Hamiltonian of the system is: | | We may seprate the Hamiltonian into three terms, <math>H=H_{x}+H_{y}+H_{z},\!</math> where |
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| | <math>H_{x}=\frac{p_{x}^{2}}{2m}+\tfrac{1}{2}m\omega^2x^2-eE_{0}x,</math> |
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| <math>H=\frac{P^2}{2m}+\frac{1}{2}m\omega ^2r^2-eE_{0}x</math> | | <math>H_{y}=\frac{p_{y}^{2}}{2m}+\tfrac{1}{2}m\omega^2y^2,</math> |
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| we seprate the Hamiltonian (<math>H=H_{x}+H_{y}+H_{z} f</math>) where
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| <math>H_{x}=\frac{p_{x}^{2}}{2m}+\frac{1}{2}m\omega ^2x^2-eE_{0}x</math>
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| <math>H_{y}=\frac{p_{y}^{2}}{2m}+\frac{1}{2}m\omega ^2y^2 </math> | | <math>H_{z}=\frac{p_{z}^{2}}{2m}+\tfrac{1}{2}m\omega^2z^2.</math> |
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| <math>H_{z}=\frac{p_{z}^{2}}{2m}+\frac{1}{2}m\omega ^2z^2</math> | | Note that each of these terms depends on only one coordinate, and that, in fact, <math>H_y\!</math> and <math>H_z\!</math> are each the Hamiltonian of a one-dimensional harmonic oscillator. In fact, if we "complete the square" in <math>H_x,\!</math> we will find that it is also a one-dimensional harmonic oscillator, but with a shifted center. Let us, in fact, do this: |
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| Notice that <math>H_{x} ,H_{z}</math>are identical to the Hamiltonian of the one dimensional harmonic oscillator, so we can write the wave function
| | <math>H_x=\frac{p_{x}^{2}}{2m}+\tfrac{1}{2}m\omega^2\left (x^2-\frac{2eE_{0}}{m\omega^2}x\right )=\frac{p_{x}^{2}}{2m}+\tfrac{1}{2}m\omega^2\left (x-\frac{eE_{0}}{m\omega^2}\right )^2-\frac{e^2E_0^2}{2m\omega^2}</math> |
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| <math>\psi (x,y,z)=\psi _{1}(x)\psi _{2}(y)\psi _{3}(z)</math>, where | | We may now easily write down the solution. If we take <math>\psi(x,y,z)=X(x)Y(y)Z(z),\!</math> then |
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| <math>\psi _{2}(y)</math>, and | | <math>X(x)=\frac{1}{2^{n_1}n_1!}\left (\frac{m\omega}{\pi\hbar}\right )^{1/4}\exp\left [-\frac{m\omega}{2\hbar}\left (x-\frac{eE_0}{m\omega^2}\right )^2\right ]H_{n_1}\left [\sqrt{\frac{m\omega}{\hbar}}\left (x-\frac{eE_0}{m\omega^2}\right )\right ],</math> |
| <math>\psi _{3}(z)</math> are the wave functions of the one dimensional harmonic oscillator:
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| <math>\psi (x,y,z)=\psi _{1}(x)\psi _{2}(y)\psi _{3}(z)</math>
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| <math>\psi _{2}(y)=\frac{1}{\sqrt{\pi \lambda 2 ^{n_{2}}n_{2}!}}H_{n_{2}}e^{\frac{-y^{2}}{2\lambda ^{2}}}</math> | | <math>Y(y)=\frac{1}{2^{n_2}n_2!}\left (\frac{m\omega}{\pi\hbar}\right )^{1/4}e^{-m\omega y^2/2\hbar}H_{n_2}\left (\sqrt{\frac{m\omega}{\hbar}}y\right ),</math> |
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| <math>\psi _{3}(z)=\frac{1}{\sqrt{\pi \lambda 2 ^{n_{3}}n_{3}!}}H_{n_{3}}e^{\frac{-z^{2}}{2\lambda ^{2}}}</math>
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| <math>\lambda =\sqrt{\frac{\hbar}{m\omega }},</math> The equation of the<math>\psi _{1}(x)</math> is
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| <math>-\frac{\hbar^{2}}{2m}\frac{\partial^2\psi _{1}(x)}{\partial x^2}+\frac{m\omega ^{2}}{2}x^{2}\psi _{1}(x)-eE_{0}(x)\psi _{1}(x)=E_{1}\psi _{1}(x)</math> | | <math>Z(z)=\frac{1}{2^{n_3}n_3!}\left (\frac{m\omega}{\pi\hbar}\right )^{1/4}e^{-m\omega z^2/2\hbar}H_{n_3}\left (\sqrt{\frac{m\omega}{\hbar}}z\right ).</math> |
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| changing variables to <math>\xi =\frac{x}{\lambda }-\frac{eE_{0}}{\sqrt{\hbar m \omega }}</math>
| | The energy may simply be written as <math>E=E_x+E_y+E_z,\!</math> where <math>E_x,\!</math> <math>E_y,\!</math> and <math>E_z\!</math> are the contributions to the energy from each of the harmonic oscillators. These are |
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| <math>\frac{\partial^2 \psi _{1}}{\partial \xi ^2}+(\frac{2E_{1}}{\hbar \omega })\frac{(eE_{0})^{2})}{\sqrt{\hbar m\omega ^{3}}})\psi _{1}-\xi ^{2}\psi _{1}=0</math> | | <math>E_x=\left (n_1+\tfrac{1}{2}\right )\hbar\omega-\frac{e^2E_0^2}{2m\omega^2},</math> |
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| we obtain the diffrential equation for a one dimensional harmonic oscillator with the solution
| | <math>E_y=\left (n_2+\tfrac{1}{2}\right )\hbar\omega,</math> |
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| <math>\psi _{1}(\xi )=\frac{1}{\sqrt{\pi \lambda 2 ^{n_{1}}n_{1}!}}H_{n_{1}}e^{\frac{-\xi ^{2}}{2\lambda ^{2}}}</math>
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| <math>\psi _{1}(\xi )=\frac{1}{\sqrt{\pi \lambda 2^{n_{1}}n_{1}!}}H_{n_{1}}(x) exp[-\frac{1}{2}(\frac{x}{\lambda }-\frac{eE_{0}}{\sqrt{\hbar m\omega ^{3}}})^{2}] | | <math>E_z=\left (n_3+\tfrac{1}{2}\right)\hbar\omega.</math> |
| (E_{1})_{n_{1}}=(n_{1}+\frac{1}{2})\hbar \omega -\frac{(eE_{0})^{2}}{2m\omega ^{2}}</math>
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| The quantization condition in this case is | | The total energy is thus |
| <math>\frac{(2E_{1})}{\hbar\omega }+\frac{(eE_{0})^2}{\hbar m\omega ^{3}}=2n_{1}+1</math>
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| so the energy eigenvalues are
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| <math>(E_{1})_{n_{1}}=(n_{1}+\frac{1}{2})\hbar \omega -\frac{(eE_{0})^{2}}{2m\omega ^{2}}</math>
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| In conclusion,the wave functions are <math>\psi (x,y,z)=\psi _{1}(x)\psi _{2}(y)\psi _{3}(z)</math>
| | <math>E=\left (n_{1}+n_{2}+n_{3}+\tfrac{3}{2}\right )\hbar\omega-\frac{e^2E_{0}^{2}}{2m\omega^{2}}.</math> |
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| <math>E_{n_{1},n_{2},n_{3}}=E_{n_{1}}+E_{n_{2}}+E_{n_{3}}=(n_{1}+n_{2}+n_{3}+\frac{3}{2})\hbar \omega -\frac{(eE_{0})^{2}}{2m\omega ^{2}}</math>
| | Back to [[Analytical Method for Solving the Simple Harmonic Oscillator#Problem|Analytical Method for Solving the Simple Harmonic Oscillator]] |
The Hamiltonian of the system is
We may seprate the Hamiltonian into three terms, 
where
and
Note that each of these terms depends on only one coordinate, and that, in fact, 
and 
are each the Hamiltonian of a one-dimensional harmonic oscillator. In fact, if we "complete the square" in 
we will find that it is also a one-dimensional harmonic oscillator, but with a shifted center. Let us, in fact, do this:
We may now easily write down the solution. If we take 
then
![{\displaystyle X(x)={\frac {1}{2^{n_{1}}n_{1}!}}\left({\frac {m\omega }{\pi \hbar }}\right)^{1/4}\exp \left[-{\frac {m\omega }{2\hbar }}\left(x-{\frac {eE_{0}}{m\omega ^{2}}}\right)^{2}\right]H_{n_{1}}\left[{\sqrt {\frac {m\omega }{\hbar }}}\left(x-{\frac {eE_{0}}{m\omega ^{2}}}\right)\right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48cf5fa6956b3d044d1c506c723a7cd1278e8268)
and
The energy may simply be written as 
where 
and 
are the contributions to the energy from each of the harmonic oscillators. These are
and
The total energy is thus
Back to Analytical Method for Solving the Simple Harmonic Oscillator